We begin by examining the first part of the formula for \(E^\star \) , which we denote for simplicity by
\(f(q) = \sum \limits _{\ell =1}^{L}\mu _\ell \sum \limits _{i \in N_\ell }^{} q^{(i)} \ln q^{(i)}.\) 
Basic calculus gives its Hessian with the entries
\(\nabla _{ii}^2f(q) = \frac{\mu _\ell }{q^{(i)}} \quad \text{for all} \; i \in N_\ell , \quad \nabla _{ij}^2f(q) = 0 \quad \text{for all} \; j \ne i. \) 
Consequently, the Hessian \(\nabla ^2 f(q) \)  is a diagonal matrix. The second order criterion for strong convexity with respect to an arbitrary norm \(\Vert \cdot \Vert \)  is given in [1]}:
\(\left\langle \nabla ^2f(q)h,h \right\rangle \ge \beta \Vert h\Vert ^2 \quad \mbox{for all } \; h \in \mathbb {R}^n.\) 
Applying this criterion in our case provides
\(\left\langle \nabla ^2f(q)h,h\right\rangle &= \sum \limits _{\ell =1}^{L}\mu _\ell \sum \limits _{i \in N_\ell }^{} \frac{\left(h^{(i)}\right)^2 }{q^{(i)}} \ge {\beta } \sum \limits _{i=1}^{n} \frac{{\left(h^{(i)}\right)}^2}{q^{(i)}} \\ & \overset{(\star )}{\ge } {\beta } \left(\sum \limits _{i=1}^n |h^{(i)}|\right)^2 = {\beta } \Vert h\Vert _1^2.\) 

The last inequality \((\star ) \)  holds due to
\(\sum \limits _{i=1}^n |h^{(i)}| = \sum \limits _{i=1}^n \frac{|h^{(i)}|}{\sqrt{q^{(i)}}}\sqrt{q^{(i)}} \le \sqrt{\sum \limits _{i=1}^{n}\frac{(h^{(i)})^2}{q^{(i)}}} \sqrt{\sum \limits _{i=1}^{n} q^{(i)}} = \sqrt{\sum \limits _{i=1}^{n}\frac{(h^{(i)})^2}{q^{(i)}}}.\) 
Taking squares on both sides of this inequality gives (\(\star \) ).
Overall, \(f\)  is \({\beta }\) -strongly convex.
Next, we turn our attention to the second part of \(E^\star \) , denoting the latter by
\(g(q)= \sum \limits _{\ell =1}^{L} (1 - \mu _\ell ) \left(\sum \limits _{i \in N_\ell }^{} q^{(i)}\right) \ln \left(\sum \limits _{i \in N_\ell }^{} q^{(i)}\right).\) 
Clearly, \(g\)  is convex in \(q\) .
It remains to recall that \(E^\star \)  – as the sum of a \(\beta \) -strongly convex function \(f\)  and the convex function \(g\)  – is \(\beta \) -strongly convex.
